Tính: 1/1+2+3 + 1/1+2+3+4 + 1/1+2+3+4+5 +……+ 1/1+2+3+……+2016 05/08/2021 Bởi Aubrey Tính: 1/1+2+3 + 1/1+2+3+4 + 1/1+2+3+4+5 +……+ 1/1+2+3+……+2016
Đáp án: $2\cdot \left(\dfrac{1}{3}-\dfrac1{2017}\right)$ Giải thích các bước giải: Ta có: $1+2+3+…+n=\dfrac{n\left(n+1\right)}{2}$ $\to \dfrac{1}{1+2+3+…+n}=\dfrac{2}{n\left(n+1\right)}=2\cdot\dfrac{1}{n\left(n+1\right)}=2\cdot\dfrac{n+1-n}{n\left(n+1\right)}$ $\to \dfrac{1}{1+2+3+…+n}=2\cdot\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)$ Mà $A=\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+\dfrac{1}{1+2+3+4+5}+…+\dfrac{1}{1+2+3+…+2016}$ $\to A=2\cdot \left(\dfrac{1}{3}-\dfrac14\right)+2\cdot \left(\dfrac{1}{4}-\dfrac15\right)+2\cdot \left(\dfrac{1}{5}-\dfrac16\right)+…+2\cdot \left(\dfrac{1}{2016}-\dfrac1{2017}\right)$ $\to A=2\cdot \left(\dfrac{1}{3}-\dfrac14+\dfrac{1}{4}-\dfrac15+\dfrac{1}{5}-\dfrac16+…+\dfrac{1}{2016}-\dfrac1{2017}\right)$ $\to A=2\cdot \left(\dfrac{1}{3}-\dfrac1{2017}\right)$ Bình luận
Đáp án: $2\cdot \left(\dfrac{1}{3}-\dfrac1{2017}\right)$
Giải thích các bước giải:
Ta có: $1+2+3+…+n=\dfrac{n\left(n+1\right)}{2}$
$\to \dfrac{1}{1+2+3+…+n}=\dfrac{2}{n\left(n+1\right)}=2\cdot\dfrac{1}{n\left(n+1\right)}=2\cdot\dfrac{n+1-n}{n\left(n+1\right)}$
$\to \dfrac{1}{1+2+3+…+n}=2\cdot\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)$
Mà $A=\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+\dfrac{1}{1+2+3+4+5}+…+\dfrac{1}{1+2+3+…+2016}$
$\to A=2\cdot \left(\dfrac{1}{3}-\dfrac14\right)+2\cdot \left(\dfrac{1}{4}-\dfrac15\right)+2\cdot \left(\dfrac{1}{5}-\dfrac16\right)+…+2\cdot \left(\dfrac{1}{2016}-\dfrac1{2017}\right)$
$\to A=2\cdot \left(\dfrac{1}{3}-\dfrac14+\dfrac{1}{4}-\dfrac15+\dfrac{1}{5}-\dfrac16+…+\dfrac{1}{2016}-\dfrac1{2017}\right)$
$\to A=2\cdot \left(\dfrac{1}{3}-\dfrac1{2017}\right)$